How do you graph ##y<=x^2+8x+16##?

Draw the parabola ##(x+4)^2 = y >=0## that has the vertex at (-4, 0) and focus at (-4, 1/4). Along with, shade the region outside the curve and its enclosure for ##y<=x^2+8x+16##.

The given inequality represents the region outside the parabola

##x^2+8x+16=(x+4)^2=y>=0##.

The vertex of this boundary is at ##(-4, 0)## and focus is at (-4, 1/4)).

For any point (x, y) on and beneath this parabola, ##y<=x^2+8x+16##.